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6 March, 17:41

What is the maximum number of grams of silver phosphate that can be produced from the reaction of 287g of silver chlorate with 52.8g of sodium phosphate?

3 Ag (ClO3) + Na3 (PO4) - > 3 Na (ClO3) + Ag3 (PO4)

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  1. 6 March, 18:04
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    The balanced equation for the above reaction is as follows;

    3AgClO₃ + Na₃PO₄ - - - > Ag₃PO₄ + 3NaClO₃

    stoichiometry of AgClO₃ to Na₃PO₄ is 3:1

    first we need to find which is the limiting reactant

    number of AgClO₃ moles - 287 g / 191.3 g/mol = 1.500 mol

    number of Na₃PO₄ moles - 52.8 g / 163.9 g/mol = 0.322 mol

    if AgClO₃ is the limiting reactant

    then 1.500 mol of AgClO₃ reacts with - 1.500/3 mol of Na₃PO₄

    therefore number of Na₃PO₄ moles required = 0.50 mol of Na₃PO₄ are required but only 0.322 mol are present

    therefore Na₃PO₄ is the limiting reactant

    amount of product formed depends on amount of limiting reactant present

    stoichiometry of Na₃PO₄ to Ag₃PO₄ is 1:1

    the number of Na₃PO₄ moles reacted - 0.322 mol

    then number of Ag₃PO₄ moles formed - 0.322 mol

    mass of Ag₃PO₄ formed - 0.322 mol x 418.6 g/mol = 134.8 g

    mass of Ag₃PO₄ produced is 134.8 g
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